(Part A)
(Part A)
Line 18: Line 18:
 
<math>h(t)=K\delta (t-a)</math>
 
<math>h(t)=K\delta (t-a)</math>
  
<math>H(s)=\int_{-\infty}^{\infty}h(\tau)e^{-s\tau}=\int_{-\infty}^{\infty}\frac{1}{2}u(\tau)e^{-s\tau}</math>
+
<math>H(s)=\int_{-\infty}^{\infty}K\delta (\tau -a)e^{-s\tau}d\tau=Ke^{-as}</math>

Revision as of 17:25, 26 September 2008

Part A

$ y(t) = K x(t-a) $

if $ x(t)=e^{jwt} $ was inputed to the system

$ y(t) = K e^{jw(t-a)} $

$ = K e^{-jwa}e^{jwt} $


eigen function is $ e^{-jwa} $


$ H(jw)=Ke^{-jwa} $

$ h(t)=K\delta (t-a) $

$ H(s)=\int_{-\infty}^{\infty}K\delta (\tau -a)e^{-s\tau}d\tau=Ke^{-as} $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn